A)
B)
C)
D)
Correct Answer: B
Solution :
[b] We know that \[\frac{dQ}{dt}=KA\frac{d\theta }{dx}\] In steady state flow of heat, \[d\theta =\frac{dQ}{dt}.\frac{1}{kA}.dx\] \[\Rightarrow \] \[{{\theta }_{H}}-\theta =k'x\Rightarrow \,\theta ={{\theta }_{H}}-K'x\] Equation \[\theta ={{\theta }_{H}}-k'x\] represents a straight line.
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